Could someone take a look at my solution to this exercise in Artin?
2.2.15. a) In the definition of subgroup, the identity element in $H$ is required to be the identity of $G$. One might require only that $H$ have an identity element, not that it is the same as the identity in $G$. Show that if $H$ has an identity at all, then it is the identity in $G$, so this definition would be equivalent to the one given.
b) Show the analogous thing for inverses.
Solution. a) Let $e$ be the identity element of $H$ and $e'$ be the identity element of $G$. Hence, \begin{align*} & \forall h \in H, \; he = eh = e \\ & \forall g \in H, \; ge' = e'g = g. \end{align*} Let $h \in H$. Then, we have $he = eh = e$ by the identity in $H$. But $H < G$, so $H \subset G$, meaning that $h \in G$, so $he' = e'h = h$ by the identity in $G$. But the identity in a group is unique, so $e = e'$. Hence, the identity in $H$ is also the identity in $G$.
(I'm not sure if a lemma is needed for uniqueness of the identity, but let me write one for the sake of completion.)
Lemma. The identity element in a group is unique.
Let $e$ and $e'$ be identity elements of $G$. Then, since $e$ is an identity, we have $$ee' = e'e = e'.$$ Since $e'$ is an identity, we have $$e'e = ee' = e.$$ Hence, $e' = e$, and the identity is unique.
b) Let $h \in H$. Then there exists an inverse $x \in H$ such that $hx = xh = e$, where $e$ is the identity in both $H$ and $G$. But $H < G$, so $H \subset G$, meaning $h \in G$. Since $G$ is a group, there exists an inverse $y \in H$ such that $hy = yh = e$. But the inverse is unique, meaning that $x = y$.
For part a, it seems to me like you showed that if $e$ is an identity in $G$, and $e$ is in $H$ a subgroup of $G$, then $e$ is an identity in $H$. Yes the identity is unique in a group, but an identity in $H$ may not be an identity in $G$. If the identity in $G$ is not itself in $H$, you have to check the possibility that some element $e'$ in $H$ is such that $he' = e'h = h$ for all $h \in H$, but that there is some $g \in G$, $g \notin H$ such that $ge' \neq g$ or $e'g \neq g$, and show that this actually can't be the case. I think you have the right idea, using uniqueness, but you need some more work (try using the fact that subgroups are closed under inversion and multiplication by another element in the subgroup). Similarly for part b.